What is the value of $\dfrac{d}{dx}\left(\sqrt{x^3}\right)$ at $x=25$ ?
Solution: The strategy We can first rewrite the radical as a rational power of $x$. Then, the derivative can be found using the power rule : $\dfrac{d}{dx}(x^n)=n\cdot x^{n-1}$ (Remember that this applies even when $n$ is a fraction.) Once we have the derivative, we can evaluate it at $x=25$. Rewriting the radical as a rational power $\sqrt{x^3}=x^{^{\frac{3}{2}}}$ Differentiating using the power rule $\begin{aligned} &\phantom{=}\dfrac{d}{dx}\left(x^{^{\frac{3}{2}}}\right) \\\\ &=\dfrac{3}{2}x^{^{\frac{3}{2}-1}} \gray{\text{The power rule}} \\\\ &=\dfrac32x^{^{\frac{1}{2}}} \end{aligned}$ Evaluating the derivative So we found that $\dfrac{d}{dx}\left(\sqrt{x^3}\right)=\dfrac32x^{^{\frac{1}{2}}}$, which can also be written as $1.5\sqrt{x}$. Now let's plug ${x=25}$ : $\begin{aligned} 1.5\sqrt{{25}}&=1.5\cdot 5 \\\\ &=7.5 \end{aligned}$ In conclusion, the value of $\dfrac{d}{dx}\left(\sqrt{x^3}\right)$ at $x=25$ is $7.5$.